f(t) = 5 e^(-0.8 t) sin(2 pi t / 3)

(b) find f'(t).

Use the product rule:

f'(t) = 5 d/dt[e^(-0.8 t)] sin(2 pi t /3) + 5 e^(-0.8 t) d/dt[sin(2 pi t / 3)]

......= 5 (-0.8) e^(-0.8) sin(2 pi t /3) + 5 e^(-0.8 t) (2 pi / 3) cos(2 pi t / 3)

......= 5 e^(-0.8 t) {(-0.8) sin(2 pi t /3) + (2 pi / 3) cos(2 pi t / 3)}

(c) Solve f'(x) = 0 to determine the maximum deflection and when it occurs.

f'(t) = 0 after cancelling the leading intrinscily non zero term gives:

(-0.8) sin(2 pi t /3) + (2 pi / 3) cos(2 pi t / 3) = 0

or:

tan( 2 pi t / 3) = 2 pi / 2.4,

so:

2 pi t / 3 =arctan(2 pi / 2.4) ~= 1.206 + pi n, n=0, 1, 2, ...

(only positive n will be considered as usually t is measured from the time

of the disturbance to the system, so times <0 are meaningless) so:

t = 0.573 + 3/2 n, n=0, 1, 2, ..

Now as e^(-0.8 t) is a strictly decreasing function the deflection of

maximum amplitude is the first of these, so it occurs at t=0.573, and is of

amplitude:

5 e^(-0.8*0.573) sin(2 pi*0.573 / 3) = 2.95

RonL